在C ++中检查一个数组是否是另一个数组的子数组

Prerequisite: std::equal() function

先决条件： std :: equal()函数

Problem statement:

问题陈述：

Check if one array is subarray of another or not.

检查一个数组是否是另一个数组的子数组。

Example:

例：

Input 1:
Arr1= [3, 4, 5, 8]
Arr2= [1, 3, 4, 5, 8, 9, 10]
Output:
True
Arr1 is subarray of arr2
Input 2:
Arr1= [3, 4, 5, 8, 9 , 10, 12]
Arr2= [1, 3, 4, 5, 8, 9, 10]
Output:
False
None is subarray of one another
Input 3:
Arr1= [3, 4, 5, 8, 9]
Arr2= [3, 4]
Output:
True
Arr2 is subarray of arr1

Solution:

解：

We already saw that we have an STL function equal() that checks whether two ranges have the same elements in order or not. We can use that function to figure out whether an array is subarray of the other one or not. No doubt, if two array sizes are not equal then the smaller size one may have the possibility of being subarray of the bigger array. If their sizes are equal then they can't be a subarray of one another anyway.

我们已经看到我们有一个STL函数equal() ，它检查两个范围是否按顺序具有相同的元素。我们可以使用该函数来确定一个数组是否是另一个数组的子数组。毫无疑问，如果两个数组大小不相等，那么较小的数组就有可能成为较大数组的子数组。如果它们的大小相等，则它们无论如何都不能成为彼此的子数组。

So there can be three cases:

因此可能有三种情况：

Both the sizes of the array are equal
数组的两个大小相等

Simple return False
简单返回False
arr1 size is greater than arr2
arr1的大小大于arr2

Check for each range of
检查每个范围

arr1 having length the same as arr2 whether equal to arr2 or not. If we can find any such range equal to arr2 then arr2 is subarray of arr1. Following is the algorithm for this case.
无论是否等于arr2， arr1的长度都与arr2相同。如果我们找到任何等于arr2的范围，则arr2是arr1的子数组。以下是这种情况的算法。
```
For i=0 to length(arr1)-length(arr2)
If equal(arr2.begin(),arr2.end(), arr1.begin()+i) is true
Then return true since we found a range 
starting from arr1[i] which is equal to arr2
Return false if no such range is found.
```
arr2 size is greater than arr1
arr2的大小大于arr1

Check for each range of
检查每个范围

arr2 having length the same as arr1 whether equal to arr1 or not. If we can find any such range equal to arr1 then arr1 is subarray of arr2. Following is the algorithm for this case.
无论是否等于arr1， arr2的长度都与arr1相同。如果我们找到任何等于arr1的范围，则arr1是arr2的子数组。以下是这种情况的算法。
```
For i=0 to length(arr2)-length(arr1)
If equal(arr1.begin(),arr1.end(), arr2.begin()+i) is true
Then return true since we found a range 
starting from arr2[i] which is equal to arr1
Return false if no such range is found.
```

C++ implantation:

C ++植入：

#include <bits/stdc++.h>
using namespace std;
void isSubarray(vector<int> arr1, vector<int> arr2)
{
if (arr1.size() == arr2.size()) {
cout << "False\n";
cout << "None of the Arrays can be subarray one other\n";
return;
}
else if (arr1.size() > arr2.size()) {
for (int i = 0; i < arr1.size() - arr2.size(); i++) {
if (equal(arr2.begin(), arr2.end(), arr1.begin() + i)) {
cout << "True\n";
cout << "Array2 is subarray of Array1\n";
}
}
}
else { //arr2.size()>arr1.size()
for (int i = 0; i < arr2.size() - arr1.size(); i++) {
if (equal(arr1.begin(), arr1.end(), arr2.begin() + i)) {
cout << "True\n";
cout << "Array1 is subarray of Array2\n";
}
}
}
}
int main()
{
cout << "Enter number of elements for array1\n";
int n;
cin >> n;
vector<int> arr1(n);
cout << "Input the elements\n";
for (int i = 0; i < n; i++)
cin >> arr1[i];
cout << "Enter number of elements for array2\n";
int m;
cin >> m;
vector<int> arr2(m);
cout << "Input the elements\n";
for (int i = 0; i < m; i++)
cin >> arr2[i];
isSubarray(arr1, arr2);
return 0;
}

Output:

输出：

Output1:
Enter number of elements for array1
4
Input the elements
3 4 5 8
Enter number of elements for array2
7
Input the elements
1 3 4 5 8 9 10
True
Array1 is subarray of Array2
Output2:
Enter number of elements for array1
7
Input the elements
3 4 5 8 9 10 12
Enter number of elements for array2
7
Input the elements
1 3 4 5 8 9 10
False
None of the Arrays can be subarray one other
Output3:
Enter number of elements for array1
5
Input the elements
3 4 5 8 9
Enter number of elements for array2
2
Input the elements
3 4
True
Array2 is subarray of Array1